Answer :
To graph the parametric equations [tex]\( x = t^2 + 2t \)[/tex] and [tex]\( y = -t \)[/tex] over the interval [tex]\( -4 \leq t \leq 1 \)[/tex], we will first consider the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] generated by [tex]\( t \)[/tex] in that range.
Let's break it down step-by-step:
1. Establish the Range of [tex]\( t \)[/tex]:
[tex]\[ -4 \leq t \leq 1 \][/tex]
2. Calculate Sample Values:
For illustrative purposes, let's calculate a few key values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
- When [tex]\( t = -4 \)[/tex]:
[tex]\[ x = (-4)^2 + 2(-4) = 16 - 8 = 8 \][/tex]
[tex]\[ y = -(-4) = 4 \][/tex]
[tex]\[ (x, y) = (8, 4) \][/tex]
- When [tex]\( t = -3 \)[/tex]:
[tex]\[ x = (-3)^2 + 2(-3) = 9 - 6 = 3 \][/tex]
[tex]\[ y = -(-3) = 3 \][/tex]
[tex]\[ (x, y) = (3, 3) \][/tex]
- When [tex]\( t = -2 \)[/tex]:
[tex]\[ x = (-2)^2 + 2(-2) = 4 - 4 = 0 \][/tex]
[tex]\[ y = -(-2) = 2 \][/tex]
[tex]\[ (x, y) = (0, 2) \][/tex]
- When [tex]\( t = -1 \)[/tex]:
[tex]\[ x = (-1)^2 + 2(-1) = 1 - 2 = -1 \][/tex]
[tex]\[ y = -(-1) = 1 \][/tex]
[tex]\[ (x, y) = (-1, 1) \][/tex]
- When [tex]\( t = 0 \)[/tex]:
[tex]\[ x = (0)^2 + 2(0) = 0 \][/tex]
[tex]\[ y = -(0) = 0 \][/tex]
[tex]\[ (x, y) = (0, 0) \][/tex]
- When [tex]\( t = 1 \)[/tex]:
[tex]\[ x = (1)^2 + 2(1) = 1 + 2 = 3 \][/tex]
[tex]\[ y = -(1) = -1 \][/tex]
[tex]\[ (x, y) = (3, -1) \][/tex]
3. Plot the Points:
- For [tex]\( t = -4 \to (8, 4) \)[/tex]
- For [tex]\( t = -3 \to (3, 3) \)[/tex]
- For [tex]\( t = -2 \to (0, 2) \)[/tex]
- For [tex]\( t = -1 \to (-1, 1) \)[/tex]
- For [tex]\( t = 0 \to (0, 0) \)[/tex]
- For [tex]\( t = 1 \to (3, -1) \)[/tex]
4. Shape of the Curve:
By plotting these points, we observe that:
- For [tex]\( t \leq -2 \)[/tex], [tex]\( x \)[/tex] starts from 8 (at [tex]\( t = -4 \)[/tex]) and decreases to 0 at [tex]\( t = -2 \)[/tex], while [tex]\( y \)[/tex] increases from 4 to 2.
- For [tex]\( -2 \leq t \leq 0 \)[/tex], [tex]\( x \)[/tex] decreases further to a minimum (negative) at [tex]\( t = -1 \)[/tex] and then rises to 0 at [tex]\( t = 0 \)[/tex], while [tex]\( y \)[/tex] continues to increase linearly from 2 to 0.
- For [tex]\( 0 \leq t \leq 1 \)[/tex], [tex]\( x \)[/tex] increases again to 3 (at [tex]\( t = 1 \)[/tex]), while [tex]\( y \)[/tex] decreases linearly from 0.
5. Conclusion:
The points and their [tex]\( x \)[/tex] and [tex]\( y \)[/tex] values indicate that the graph will start at [tex]\( (8, 4) \)[/tex] when [tex]\( t = -4 \)[/tex], move towards [tex]\( (3, 3) \)[/tex], [tex]\( (0, 2) \)[/tex], [tex]\( (-1, 1) \)[/tex], [tex]\( (0, 0) \)[/tex], and finally end at [tex]\( (3, -1) \)[/tex] when [tex]\( t = 1 \)[/tex].
The resulting graph will resemble an arc or a curve with vertices moving as described through the parametric equations, passing through the points we calculated. It starts high and right, loops down and to the left, turns up as it moves leftward, and finally curves back up and right.
Let's break it down step-by-step:
1. Establish the Range of [tex]\( t \)[/tex]:
[tex]\[ -4 \leq t \leq 1 \][/tex]
2. Calculate Sample Values:
For illustrative purposes, let's calculate a few key values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
- When [tex]\( t = -4 \)[/tex]:
[tex]\[ x = (-4)^2 + 2(-4) = 16 - 8 = 8 \][/tex]
[tex]\[ y = -(-4) = 4 \][/tex]
[tex]\[ (x, y) = (8, 4) \][/tex]
- When [tex]\( t = -3 \)[/tex]:
[tex]\[ x = (-3)^2 + 2(-3) = 9 - 6 = 3 \][/tex]
[tex]\[ y = -(-3) = 3 \][/tex]
[tex]\[ (x, y) = (3, 3) \][/tex]
- When [tex]\( t = -2 \)[/tex]:
[tex]\[ x = (-2)^2 + 2(-2) = 4 - 4 = 0 \][/tex]
[tex]\[ y = -(-2) = 2 \][/tex]
[tex]\[ (x, y) = (0, 2) \][/tex]
- When [tex]\( t = -1 \)[/tex]:
[tex]\[ x = (-1)^2 + 2(-1) = 1 - 2 = -1 \][/tex]
[tex]\[ y = -(-1) = 1 \][/tex]
[tex]\[ (x, y) = (-1, 1) \][/tex]
- When [tex]\( t = 0 \)[/tex]:
[tex]\[ x = (0)^2 + 2(0) = 0 \][/tex]
[tex]\[ y = -(0) = 0 \][/tex]
[tex]\[ (x, y) = (0, 0) \][/tex]
- When [tex]\( t = 1 \)[/tex]:
[tex]\[ x = (1)^2 + 2(1) = 1 + 2 = 3 \][/tex]
[tex]\[ y = -(1) = -1 \][/tex]
[tex]\[ (x, y) = (3, -1) \][/tex]
3. Plot the Points:
- For [tex]\( t = -4 \to (8, 4) \)[/tex]
- For [tex]\( t = -3 \to (3, 3) \)[/tex]
- For [tex]\( t = -2 \to (0, 2) \)[/tex]
- For [tex]\( t = -1 \to (-1, 1) \)[/tex]
- For [tex]\( t = 0 \to (0, 0) \)[/tex]
- For [tex]\( t = 1 \to (3, -1) \)[/tex]
4. Shape of the Curve:
By plotting these points, we observe that:
- For [tex]\( t \leq -2 \)[/tex], [tex]\( x \)[/tex] starts from 8 (at [tex]\( t = -4 \)[/tex]) and decreases to 0 at [tex]\( t = -2 \)[/tex], while [tex]\( y \)[/tex] increases from 4 to 2.
- For [tex]\( -2 \leq t \leq 0 \)[/tex], [tex]\( x \)[/tex] decreases further to a minimum (negative) at [tex]\( t = -1 \)[/tex] and then rises to 0 at [tex]\( t = 0 \)[/tex], while [tex]\( y \)[/tex] continues to increase linearly from 2 to 0.
- For [tex]\( 0 \leq t \leq 1 \)[/tex], [tex]\( x \)[/tex] increases again to 3 (at [tex]\( t = 1 \)[/tex]), while [tex]\( y \)[/tex] decreases linearly from 0.
5. Conclusion:
The points and their [tex]\( x \)[/tex] and [tex]\( y \)[/tex] values indicate that the graph will start at [tex]\( (8, 4) \)[/tex] when [tex]\( t = -4 \)[/tex], move towards [tex]\( (3, 3) \)[/tex], [tex]\( (0, 2) \)[/tex], [tex]\( (-1, 1) \)[/tex], [tex]\( (0, 0) \)[/tex], and finally end at [tex]\( (3, -1) \)[/tex] when [tex]\( t = 1 \)[/tex].
The resulting graph will resemble an arc or a curve with vertices moving as described through the parametric equations, passing through the points we calculated. It starts high and right, loops down and to the left, turns up as it moves leftward, and finally curves back up and right.
